# Predicting amounts of reaction products

What volume of $\ce{H2O(g)}$ is produced when 7.60 mol of $\ce{C2H4(g)}$ reacts at STP?

The reaction is $\ce{C2H4 + 3O2} = \ce{2CO2 + 2H2O}$.

Stoichiometry: 7.6 mol of $\ce{C2H4}$, for every 1 mol of $\ce{C2H4}$ there's 2 mol of $\ce{H2O}$ so $7.6\cdot2 = 15.2$ mol of $\ce{H2O}$. At STP 1 mol of any gas = 22.4 L, $15.2\cdot22.4 = 340.40$ L of $\ce{H2O}$.

Did I do it correctly, if not where did I go wrong and what method should I try instead of the one I did?

• Welcome to Chemistry Stack Exchange. Your working out looks correct to me and that should be the correct answer – Nanoputian Oct 3 '15 at 6:03
• Please note that, according to current IUPAC recommendations, STP corresponds to a temperature of $T=273.15\ \mathrm K$ and a pressure of $p=100\,000\ \mathrm{Pa}$. At this temperature and pressure, the molar volume of an ideal gas is not $V_\mathrm m=22.4\ \mathrm{l/mol}$; it actually is $V_\mathrm m=22.710\,947(13)\ \mathrm{l/mol}$. – Loong Oct 3 '15 at 9:53
• Please make titles more descriptive so that others can benefit from your question. – jerepierre Oct 3 '15 at 10:38

## 1 Answer

You have done it correctly.

I have one small point: You should use the correct number of significant figures. You know the number of moles in three significant figures and the conditions at STP in at least three significant figures, so you should give the number of liters of water in three significant figures also. You may want to add a sentence like: Thus there are 340 liters of water.